Transcript Math80A-Lecture2-3
Slide 1
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 2
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 3
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 4
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 5
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 6
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 7
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 2
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 3
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 4
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 5
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 6
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH
Slide 7
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS
Investment Problems
Example 1 An investment counselor invested 75% of a client’s money into a 9%
annual simple interest money market fund. The remainder was invested in 6%
annual simple interest government securities. Find the amount invested in each if
the total annual interest earned is $3300. Annual Interest is the interest earned
EACH YEAR (so t=1).
Step 1) What are we being asked to find?
The amount invested in each account.
Let x be the total amount invested, then
75% of x is the amount invested in the Money Market Fund and
25% of x is the amount invested in the government securities.
Given info:
P
r
t
I
Money
Market Fund
.75x
.09
1
.75x(.09)
Gov.
securities
.25x
.06
1
.25x(.06)
TOTAL
$3300
Step 2) Form an equation to solve for the unknown quantity.
.75x(.09) + .25x(.06) = $3300
Step 3) Solve the equation:
.0675x + .015x = $3300
.0825x = $3300
x = $40,000
This column
can be deleted
since
t just equals 1.
Step 4) Check the result.
.75($40,000)(.09) + .25($40,000)(.06) = $3300?
Step 5) State the conclusion. We were asked to find amt in each account.
75% of x is the amount invested in the Money Market Fund = .75($40,000) = $30,000
25% of x is the amount invested in the government securities = .25($40,000) = $10,000
$30,000 invested in Money Market Fund and $10,000 invested in gov. securities.
You try this one:
An investment of $2500 is made at an annual simple interest
rate of 7%. How much additional money must be invested at
10% so that the total interest earned will be 9% of the total
investment?
Step 1) What are we being asked to find?
The additional money to be invested at 10% to earn total interest
that is 9% of the total investment.
Let x = additional money to be invested.
Given info: : Original investment = $2500
annual simple interest rate = 7%
x = additional investment
10% = new rate
New Interest = 9% of TOTAL investment
TOTAL investment = 2500 + x
P
r
I
Original Investment
at 7%
2500
.07
.07(2500)=
d
175
Additional
Investment at 10%
x
.10
.10x d
TOTAL
2500 + x
.09d
.09(2500+x)
d
Step 2: Form an equation 175 + .10x = .09(2500 + x)
Step 3: Solve the equation 175 + .10x = 225 + .09x
.01x = 50
x = 5000
Step 4: Check the result 175 + .10(5000) = .09(2500 + 5000? Yes
Step 5: State the conclusion $5000 should be invested at 10%_
Value Mixture Problems
Strategy:
Make a table with a row for each ingredient in the mixture. These
ingredients are listed in the first column. For each ingredient, write a
numerical or variable expression for the amount of the ingredient
used, the unit cost of the bend, and the value of the amount used. The
last row of the table will be for the resulting blend (or mixture) of the
ingredients. In the last row you ADD the values in the previous rows,
EXCEPT for the UNIT COST. The new unit cost will be GIVEN to
you or it will be the quantity you are solving for. DO NOT ADD
UNIT COSTS.
You then use the equation, Amount*Unit Cost = Value
to solve for the unknown quantity
Amount = how much is used (total ounces, pounds, etc. of each
ingredient.
Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost
per pound, $/lb)
Value = amount in dollars of the total amount
Amount * Unit Cost = Value
Example: 10 oz * $8/oz = $80
Example:
How many ounces of silver alloy that costs $6 per ounce must be mixed
with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that
costs $6.50 per ounce?
What are we being asked to find? How many ounces (AMOUNT) of the
$6 alloy.
Let x = amount in ounces of $6 alloy
AMOUNT (oz)
UNIT COST ($/oz)
VALUE($)
$6 per oz Alloy
x
6
6x
$8 per oz Alloy
10
8
80
$6.50 per Alloy
(mixture)
x + 10
6.50
$6x + $80
Equation to solve for x:
(x + 10)($6.50) = $6x + $80
$6.50x + $65 = $6x + $80
$6.50x - $6x = $80 - $65
$0.50x = $15
x = 30
Check: Does (30 + 10)(6.50) = 6(30) + 80?
(40)(6.50)
dd = 180 + 80
260 = 260 Yes
State Conclusion: What does x represent? The number of ounces of $6 alloy
CONCLUSION: 30 OUNCES NEEDdd
TO BE ADDED.
PERCENT MIXTURE PROBLEMS
The strategy for these problems is virtually the same as for the value mixture
problems, but you use a different equation.
Recall the Percent Triangle:
Amount = Percent * Base
In these problems, the amount
of each ingredient will actually
be the “Base” in our equation.
The Amount will be the
Quantity of substance in
each ingredient. So the new formula is
Amt
(QUANTITY
Of SUBSTANCE)
Percent
Base
(AMOUNT of
INGREDIENT)
Quantity = Percent * Amount
Example
A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid
solution and a 4% acid solution. How many liters of each solution should
the chemist use?
Step 1:
What are we being asked to find? How many liters (AMOUNT) of each solution
that the chemist should use to make 3L (amount of mixture) of a 7% (percent
acid of mixture) acid solution.
Given: The mixture will contain 3L of 7% acid.
One ingredient is 9% acid and the other is 4% acid.
If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll
use variables.
Let x= amount of 9% acid solution.
Since the total amount is 3L
3 – x = amount of 4% solution.
PERCENT
(changed to
decimal)
AMOUNT (L)
QUANTITY
(L)
9% acid
.09
xd
.09x
d
4% acid
.04
3 –d x
.04(3-x)
d
7 % acid
(mixture)
.07d
3d
.07(3)
d
Notice in this table that if we use the last row for our equation,
there is no x to solve for:
.07(3) = .07(3)
However, if we look at the last column on the left, we can
make an equation. We can add the quantities together to get the quantity of acid
in the final mixture.
.09x + .04(3-x)
d = .07(3)
.09x + .12 -d.04x = .21
.09x - .04xdd
= .21 - .12
.05x = .09 dd
x = .09/.05dd
x = 1.8 dd
Check: Does .09(1.8) + .04(3-1.8) = .07(3) ?
.162 + .04(1.2) = .21
dd = .21
.162 + .048
.21 = .21 YES
State the Conclusion. What were we being asked to find?
The amount of each acid solution, 9% and 4%.
x = amount of 9% acid solution in liters = 1.8 L
3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L
CONCLUSION:
MIX 1.8 L OF 9% ACID SOLUTIONdd
AND 1.2 L OF 4% ACID SOLUTION
CH. 2.3 UNIFORM MOTION PROBLEMS
Just as we did not add unit costs or percentages in
Mixture problems, we also DO NOT ADD RATES
in uniform motion problems. The only time
we add rates is when an object’s rate is affected
by another force such as the wind or a current.
Also, we do not add TIMES if the objects traveled
at the same time, but we do if they traveled at
different times. Remember, it sometimes helps
to draw pictures with these problems.
Distance
Rate
(speed)
Time
Example 1 p.223:
Two cars, the first traveling 10 mph FASTER THAN the second, start at the same
time from the same point and travel in opposite directions. In 3 hours they are 288
miles apart. Find the rate of the second car.
Step 1) Analyze the problem. What are we asked to find? The rate of the second car.
If we let x= rate of the second car, and the first car is 10 mph faster than the second
car, then x + 10 will be the rate of the first car.
Draw a picture:
10 + x mph
x mph
Total Distance = 288 miles Time =3 hours
Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance +
Car 2 distance, since they went in opposite directions from the same point, and we are
given that total distance is 288 miles.
Rate
Time
Distance
Car 1
x + 10
3
3(x + 10)
Car 2
x
3
3x
TOTAL
288
Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288
3x + 30 + 3x = 288
6x = 258
x = 43
CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH